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• playing an mp3 into a voice channel with discord.py

  12 décembre 2020, par Tanuj KS

  I'm trying to make a text-to-speech bot for people to use in no-mic-chats to speak into voice channels. I found other articles explaining this but most of them show the windows ffmpeg version but I am on a Mac. this is my code so far :

  


  @bot.command()
async def speak(ctx, message):
    tts = gtts.gTTS(message, lang="en")
    tts.save("text.mp3")
    if ctx.guild.voice_client:
        vc = ctx.guild.voice_client
    else:
        voice_channel = get(ctx.guild.voice_channels, name="Voice Lounge")
        vc = await voice_channel.connect()
    vc.play(source='text.mp3')


  


  This gives me the error :
discord.ext.commands.errors.CommandInvokeError: Command raised an exception: ClientException: ffmpeg was not found.

  


  Some people said to specify the FFMPEG.exe file but I do not see that. I downloaded the FFMPEG git from git clone https://git.ffmpeg.org/ffmpeg.git ffmpeg

  


  but I still get the error. other articles said to specify the executable but I can't find any ffmpeg.exe which I have to specify the path for.

  


  Thanks in advance

  


• Capture FFMPEG output in PHP

  22 novembre 2012, par Andrew

  I need to read the output from ffmpeg in order to even try the solution to my question from yesterday. This is a separate issue from my problem there, so I made a new question.

  How the heck do I get the output from an ffmpeg -i command in PHP ?

  This is what I've been trying :

  <?PHP
  
      error_reporting(E_ALL);
  
      $src = "/var/videos/video1.wmv";
  
      $command = "/usr/bin/ffmpeg -i " . $src;
  
      echo "<b>",$command,"</b><br />";
  
      $command = escapeshellcmd($command);
  
  
  
      echo "backtick:<br /><pre>";
  
      `$command`;
  
  
  
      echo "</pre><br />system:<br /><pre>";
  
      echo system($command);
  
  
  
      echo "</pre><br />shell_exec:<br /><pre>";
  
      echo shell_exec($command);
  
  
  
      echo "</pre><br />passthru:<br /><pre>";
  
      passthru($command);
  
  
  
      echo "</pre><br />exec:<br /><pre>";
  
      $output = array();
  
      exec($command,$output,$status);
  
      foreach($output AS $o)
  
      {
  
              echo $o , "<br />";
  
      }
  
      echo "</pre><br />popen:<br /><pre>";
  
      $handle = popen($command,&#39;r&#39;);
  
      echo fread($handle,1048576);
  
      pclose($handle);
  
      echo "</pre><br />";
  
  ?>

  This is my output :

  <b>/usr/bin/ffmpeg -i /var/videos/video1.wmv</b><br />
  
  backtick:<br />
  
      <pre></pre><br />
  
  system:<br />
  
      <pre></pre><br />
  
  shell_exec:<br />
  
      <pre></pre><br />
  
  passthru:<br />
  
      <pre></pre><br />
  
  exec:<br />
  
      <pre></pre><br />
  
  popen:<br />
  
      <pre></pre><br />

  I don't get it. safe_mode is off. There's nothing in disable_functions. The directory is owned by www-data (the apache user on my Ubuntu system). I get a valid status back from exec() and system() and running the same command from the command line give me tons of output. I feel like I must be missing something obvious but I have no idea what it is.

  Help. Please.

• Capture FFMPEG output

  2 mai 2017, par Andrew

  I need to read the output from ffmpeg in order to even try the solution to my question from yesterday. This is a separate issue from my problem there, so I made a new question.

  How the heck do I get the output from an ffmpeg -i command in PHP ?

  This is what I’ve been trying :

  <?PHP
  
      error_reporting(E_ALL);
  
      $src = "/var/videos/video1.wmv";
  
      $command = "/usr/bin/ffmpeg -i " . $src;
  
      echo "<b>",$command,"</b><br />";
  
      $command = escapeshellcmd($command);
  
  
  
      echo "backtick:<br /><pre>";
  
      `$command`;
  
  
  
      echo "</pre><br />system:<br /><pre>";
  
      echo system($command);
  
  
  
      echo "</pre><br />shell_exec:<br /><pre>";
  
      echo shell_exec($command);
  
  
  
      echo "</pre><br />passthru:<br /><pre>";
  
      passthru($command);
  
  
  
      echo "</pre><br />exec:<br /><pre>";
  
      $output = array();
  
      exec($command,$output,$status);
  
      foreach($output AS $o)
  
      {
  
              echo $o , "<br />";
  
      }
  
      echo "</pre><br />popen:<br /><pre>";
  
      $handle = popen($command,'r');
  
      echo fread($handle,1048576);
  
      pclose($handle);
  
      echo "</pre><br />";
  
  ?>

  This is my output :

  <b>/usr/bin/ffmpeg -i /var/videos/video1.wmv</b><br />
  
  backtick:<br />
  
      <pre></pre><br />
  
  system:<br />
  
      <pre></pre><br />
  
  shell_exec:<br />
  
      <pre></pre><br />
  
  passthru:<br />
  
      <pre></pre><br />
  
  exec:<br />
  
      <pre></pre><br />
  
  popen:<br />
  
      <pre></pre><br />

  I don’t get it. safe_mode is off. There’s nothing in disable_functions. The directory is owned by www-data (the apache user on my Ubuntu system). I get a valid status back from exec() and system() and running the same command from the command line give me tons of output. I feel like I must be missing something obvious but I have no idea what it is.